\(CH_4+ZO_2\rightarrow CO_2+2H_2O\)

0,2         0,4        0,2

\(n_{CH_4}=\dfrac{m}{M}=\dfrac{3,2}{16}=0,2\left(mol\right)\)

\(\Rightarrow V_{O_2}=8,96l\\ m_{CO_2}=n.M=0,2.44=8,8\left(g\right)\)

\(\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)

\(=\dfrac{1.2+1.2.2^2+1.2.3^2+1.2.4^2+1.2.5^2}{3.4+3.4.2^2+3.4.3^2+3.4.4^2+3.4.5^2}\)

\(=\dfrac{1.2\left(1+2^2+3^2+4^2+5^2\right)}{3.4\left(1+2^2+3^2+4^2+5^2\right)}\\ =\dfrac{2}{12}=\dfrac{1}{6}\)

a)\(H_3PO_4:\)

các nguyên tố:\(Hidrô,phospho,oxi\)

tỉ lệ:\(3:1:4\)

p tử khối :98

b)\(CaCO_3\)

các nguyên tố:\(canxi,cacbon,oxi\)

tỉ lệ:1:1:3

p tử khối:100

c)\(Fe_2O_3\)

các nguyên tố:\(sắt,oxi\)

tỉ lệ:2:3

p tử khối:160

\(3x\left(4x-3\right)-\left(2x-1\right)\left(6x+5\right)\\ =12x^2-9x-\left(12x^2+10x-6x-5\right)\)

\(=12x^2-9x-\left(12x^2+4x-5\right)\\ =5-13x\)

\(I_2=0,75A\\ R_2=\dfrac{U_2}{I_2}=\dfrac{30}{0,75}=40\text{Ω}\)

\(R_1=\dfrac{R_2}{2}=\dfrac{40}{2}=20\text{Ω}\)

\(U_1=I_1.R_1=0,5.20=10V\)

a)\(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)

\(=\left[\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]:\left(\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\right)\)

\(=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}-1}.\dfrac{1}{x-1}\)

b)\(x=5+2\sqrt{3}\Rightarrow A=\dfrac{1}{5+2\sqrt{3}-1}=\dfrac{1}{4+2\sqrt{3}}=\dfrac{1}{3+2\sqrt{3}+1}\)

\(\Leftrightarrow A=\dfrac{1}{\left(\sqrt{3}+1\right)^2}\Rightarrow\sqrt{A}=\dfrac{1}{\sqrt{3}+1}\)

đề tham khảo đi bạn!!!!!

lần trước bị nhắc nhở rồi mà còn tái phạm 

xếp thành hình ngôi sao 5 cánh nhé!!!

cop nhớ đè tham khảo:https://lazi.vn/edu/exercise/co-5-dung-dich-01m-dung-trong-5-lo-mat-nhan-na2co3-baoh2-naoh-khso4-kcl-neu-khong-dung không thì bị CTV xóa câu trả lời đấy